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Part 1A Simple Approach
ToShort Circuit
Calculations
Bulletin EDP1(20041)
EngineeringDependableProtectionFor AnElectricalDistributionSystem
Bussmann

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Electrical Distribution System
Basic Considerations of ShortCircuit Calculations
Sources of short circuit current that are normally taken
under consideration include:
 Utility Generation
 Local Generation
 Synchronous Motors and Induction Motors
Capacitor discharge currents can normally be
neglected due to their short time duration. Certain IEEE
(Institute of Electrical and Electronic Engineers) publications
detail how to calculate these currents if they are substantial.
Asymmetrical ComponentsShort circuit current normally takes on an asymmetrical
characteristic during the first few cycles of duration. That is,
it is offset about the zero axis, as indicated in Figure 1.
Figure 1
In Figure 2, note that the total short circuit current Ia is
the summation of two components  the symmetrical RMS
current IS, and the DC component, IDC. The DC componentis a function of the stored energy within the system at the
initiation of the short circuit. It decays to zero after a few
cycles due to I2R losses in the system, at which point the
short circuit current is symmetrical about the zero axis. The
RMS value of the symmetrical component may be deter
mined using Ohm`s Law. To determine the asymmetrical
component, it is necessary to know the X/R ratio of the
system. To obtain the X/R ratio, the total resistance and total
reactance of the circuit to the point of fault must be
determined. Maximum thermal and mechanical stress on
the equipment occurs during these first few cycles. It is
important to concentrate on what happens dur ing the first
half cycle after the initiation of the fault.
3
Why ShortCircuit CalculationsSeveral sections of the National Electrical Code relate
to proper overcurrent protection. Safe and reliable
application of overcurrent protective devices based on
these sections mandate that a short circuit study and aselective coordination study be conducted.
These sections include, among others:
1109 Interrupting Rating
11010 Component Protection
23065 Service Entrance Equipment
2401 Conductor Protection
25095 Equipment Grounding Conductor Protection
51717 Health Care Facilities  Selective Coordination
Compliance with these code sections can best be
accomplished by conducting a shor t circuit study and a
selective coordination study.
The protection for an electrical system should not onlybe safe under all service conditions but, to insure continuity
of service, it should be selectively coordinated as well. A
coordinated system is one where only the faulted circuit is
isolated without disturbing any other part of the system.
Overcurrent protection devices should also provide short
circuit as well as overload protection for system
components, such as bus, wire, motor controllers, etc.
To obtain reliable, coordinated operation and assure
that system components are protected from damage, it is
necessary to first calculate the available fault current at
various critical points in the electrical system.
Once the shortcircuit levels are determined, the
engineer can specify proper interrupting rating require
ments, selectively coordinate the system and providecomponent protection.
General Comments on ShortCircuit CalculationsShort Circuit Calculations should be done at all critical
points in the system.
These would include:
 Service Entrance
 Panel Boards
 Motor Control Centers
 Motor Starters
 Transfer Switches
 Load Centers
Normally, short circuit studies involve calculating abolted 3phase fault condition. This can be characterized
as all three phases bolted together to create a zero
impedance connection. This establishes a worst case
condition, that results in maximum thermal and mechanical
stress in the system. From this calculation, other types of
fault conditions can be obtained.
C
U
R
R
E
N
T
TIME

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3 ShortCircuit Current Calculations
Procedures and Methods
3 ShortCircuit Current Calculations,Procedures and Methods
To determine the fault current at any point in the
system, first draw a oneline diagram showing all of the
sources of shortcircuit current feeding into the fault, as well
as the impedances of the circuit components.To begin the study, the system components, including
those of the utility system, are represented as impedances
in the diagram.
The impedance tables given in the Data Section
include three phase and single phase transformers, current
transformers, safety switches, circuit breakers, cable, and
busway. These tables can be used if information from the
manufacturers is not readily available.
It must be understood that short circuit calculations are
performed without current limiting devices in the system.
Calculations are done as though these devices are
replaced with copper bars, to determine the maximum
available short circuit current. This is necessary to
project how the system and the current limiting devices willperform.
Also, current limiting devices do not operate in series
to produce a compounding current limiting effect. The
downstream, or load side, fuse will operate alone under a
short circuit condition if properly coordinated.
5
System A3 Single Transformer System
M
Available UtilityS.C. MVA 100,000
1500 KVA Transformer480Y/277V,3.5%Z, 3.45%X, .56%RIf.l. = 1804A
25  500kcmil6 Per PhaseService Entrance Conductorsin Steel Conduit
2000A Switch
KRPC2000SP FuseMain Swbd.
400A Switch
LPSRK400SP Fuse
50  500 kcmilFeeder Cable in Steel Conduit
Fault X2MCC No. 1
Motor
Fault X1
System B3 Double Transformer System
1000 KVA Transformer,
480/277 Volts 33.45%X, .60%RIf.l. = 1203A
Available UtilityS.C. KVA 500,000
30  500 kcmil4 Per Phase
Copper in PVC Conduit 1600A Switch
KRPC1500SP Fuse
LPSRK350SP Fuse
400A Switch
225 KVA208/120 Volts 3.998%X, .666%R
20  2/02 Per PhaseCopper in PVC Conduit
In this example, assume 0% motor load.
2
To begin the analysis, consider the following system,
supplied by a 1500 KVA, three phase transformer having a
full load current of 1804 amperes at 480 volts. (See System
A, below) Also, System B, for a double transformation, will
be studied.To start, obtain the available shortcircuit KVA, MVA, or
SCA from the local utility company.
The utility estimates that System A can deliver a short
circuit of 100,000 MVA at the primary of the transformer.
System B can deliver a shortcircuit of 500,000 KVA at the
primary of the first transformer. Since the X /R ratio of the
utility system is usually quite high, only the reactance need
be considered.
With this available shortcircuit information, begin to
make the necessary calculations to determine the fault
current at any point in the electrical system.
Four basic methods will be presented in this text to
instruct the reader on short circuit calculations.
These include : the ohmic method
 the per unit method
 the TRON Computer Software method
 the point to point method
1
2
Note: The above 1500KVA transformer serves 100% motor load.
Fault X1
Fault X2
1

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3 ShortCircuit Current Calculations Procedures and Methods
Ohmic Method
6
3 Short Circuit Calculations,Ohmic Method
Most circuit component impedances are given in ohms
except utility and transformer impedances which are found
by the following formulae* (Note that the transformer andutility ohms are referred to the secondary KV by squaring
the secondary voltage.)
Step 1. Xutility =1000 (KVsecondary)2
S.C. KVAutility
Step 2. Xtrans =(10)(%X**)(KVsecondary)2
KVAtrans
Rtrans =(10)(%R**)(KVsecondary)2
KVAtrans
Step 3. The impedance (in ohms) given for currenttransformers, large switches and large circuit breakers isessentially all X.
Step 4. Xcable and bus .Rcable and bus.
Step 5. Total all X and all R in system to point of fault.
Step 6. Determine impedance (in ohms) of the system by:
ZT = (RT)2 + (XT)2
Step 7. Calculate shortcircuit symmetrical RMS amperes
at the point of fault.
IS.C. sym RMS=Esecondary lineline
3 (ZT)
Step 8. Determine the motor load. Add up the full loadmotor currents. The full load motor current in the system is
generally a percentage of the transformer full load current,
depending upon the types of loads. The generally
accepted procedure assumes 50% motor load when both
motor and lighting loads are considered, such as supplied
by 4 wire, 208Y/120V and 480Y/277V volt 3phase
systems.)
*For simplicity of calculations all ohmic values are single phase distance one way, later compensated for in the three phase shortcircuit formula by the factor, 3.(See Step 7.)
**UL Listed transformers 25 KVA and larger have a 10% impedance tolerance. Short circuit amperes can be affected by this tolerance.Only X is considered in this procedure since utility X/R ratios are usually quite high. For more finite details obtain R of utility source.A more exact determination depends upon the subtransient reactance of the motors in question and associated circuit impedances. A less conservativemethod would involve the total motor circuit impedance to a common bus (sometimes referred to as a zero reactance bus).
Arithmetical addition results in conservative values of fault current. More finite values involve vectorial addition of the currents.
Note: The ohms of the circuit components must be referred to the same voltage. If there is more than one voltage transformation in the system, the ohmicmethod becomes more complicated. It is recommended that the perunit method be used for ease in calculation when more than one voltage transformationexists in the system.
Step 9. The symmetrical motor contribution can beapproximated by using an average multiplying factor
associated with the motors in the system. This factor varies
according to motor design and in this text may be chosen
as 4 times motor full load current for approximatecalculation purposes. To solve for the symmetrical motor
contribution:
Isym motor contrib = (4) x (Ifull load motor)
Step 10. The total symmetrical shortcircuit RMS current iscalculated as:
Itotal S.C. sym RMS = (IS.C. sym RMS ) + (Isym motor contrib)
Step 11. Determine X/R ratio of the system to the point offault.
X/Rratio =Xtotal Rtotal
Step 12. The asymmetrical factor corresponding to the X/Rratio in Step 11 is found in Table 8, Column Mm. This
multiplier will provide the worst case asymmetry occurring
in the first 1/2 cycle. When the average 3phase multiplier
is desired use column Ma.
Step 13. Calculate the asymmetrical RMS shortcircu itcurrent.
IS.C. asym RMS = (IS.C. sym RMS) x (Asym Factor)
Step 14. The shortcircuit current that the motor load cancontribute is an asymmetrical current usually approximated
as being equal to the locked rotor current of the motor.
As a close approximation with a margin of safety use:
Iasym motor contrib = (5) x (Ifull load motor)
Step 15. The total asymmetrical shortcircuit RMS current iscalculated as:
Itotal S.C. asym RMS = (IS.C. asym RMS) + (Iasym motor contrib)

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M M
3 ShortCircuit Current Calculations Procedures and Methods
Ohmic Method To Fault X1 System A
Available UtilityS.C. MVA 100,000
1500 KVA Transformer,480V, 3,3.5%Z, 3.45%X , 0.56%R
If.l. trans = 1804A
25  500 kcmil6 Per PhaseService EntranceConductors in Steel Conduit
2000A Switch
KRPC2000SP Fuse
Fault X1
Motor Contribution
7
R X
X =1000(.48)2
= 0.0000023 0.0000023100,000,000
X =(10) (3.45) (.48)2
= 0.0053 0.00531500
R =(10) (.56) (.48)2
= 0.00086 0.00086 1500
X =25'
x0.0379
= 0.000158 0.0001581000 6
R = 25' x 0.0244 = 0.000102 0.000102 1000 6
X = 0.000050 0.000050
Total R and X = 0.000962 0.00551
Ztotal per = (0.000962)2 + (0.00551)2 = 0.0056phase
IS.C. sym RMS =480 = 49,489A
3 (.0056)
Isym motor contrib = 4 x 1804 = 7216A(100% motor load)
Itotal S.C. sym RMS = 49,489 + 7216 = 56,705A(fault X1)
X/Rratio = .00551 = 5.73.000962
Asym Factor = 1.294 (Table 8)
IS.C. asym RMS = 1.294 x 49,489 = 64,039A
Iasym motor contrib = 5 x 1804 = 9,020A(100% motor load)
Itotal S.C. asym RMS = 64,039 + 9,020 = 73,059A(fault X1)
OneLine D iagram Impedance D iagram
1 1
Note: See Ohmic Method Procedure for Formulas.
(Table 1.2)
(Table 3)
(Table 5)

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R X
X = 0.00551 0.00551
R = 0.000962 0.000962
X = .00008 0.00008
X =50 x .0379 = 0.00189 0.00189
1000
R =50 x .0244 = 0.00122 0.00122
1000
Total R and X = 0.002182 0.00748
Ztotal per = (0.002182)2 + (0.00748)2 = 0.00778phase
IS.C. sym RMS =480 = 35,621A
3 (.00778)
Isym motor contrib = 4 x 1804 = 7216A(100% motor load)
Itotal S.C. sym RMS = 35,621 + 7,216 = 42,837A(fault X2)
X/Rratio =.00748 = 3.43
.002182
Asym Factor=
1.149 (Table 8)IS.C. asym RMS = 1.149 x 35,621 = 40,929A
Iasym motor contrib = 5 x 1804 = 9,020A(100% motor load)
Itotal S.C. asym RMS = 40,929 + 9,020 = 49,949A(fault X2)
Note: See Ohmic Method Procedure for Formulas. Actual motor contributionwill be somewhat smaller than calculated due to the impedance of thefeeder cable.
8
3 ShortCircuit Current Calculations Procedures and Methods
Ohmic Method To Fault X2 System A
M M
Adjusted Impedanceto Fault X1
Fault X1
400A Switch
LPSRK400SP Fuse
50  500 kcmilFeeder Cablein Steel Conduit
Fault X2
Motor Contribution
OneLine Diagram Impedance Diagram
2 2
1 1
(Table 3)
(Table 5)

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To use the OHMIC Method through a second transformer,
the following steps apply:
Step 1a. Summarize X and R values of all components onprimary side of transformer.
OneLine Diagram Impedance Diagram
Step 1b. Reflect X and R values of all components tosecondary side of transformer
Xs =Vs2 (Xp) Rs =
Vs2 (Rp)Vp2 Vp2
and proceed with steps 2 thru 15 from page 6.
3 ShortCircuit Current Calculations Procedures and Methods
Ohmic Method To Fault X1 System B
9
R X
X =1000 (.48)2
= .000461 .000461500,000
X =(10) (3.45) (.48)2
= .00795 .007951000
R =(10) (.60) (.48)2
= .00138 .00138 1000
X =30'
x.0303 = .000227 .000227
1000 4
R =30'
x.0220
= .000165 .000165 1000 4
X = .000050 .00005
Total R and X = .001545 .008688
Ztotal per = (.001545)2 + (.008688)2 = .008824phase
IS.C. sym RMS =480 = 31,405A
3 (.008824)
X/Rratio =.008688 = 5.62.001545
Asym Factor = 1.285 (Table 8)
IS.C. asym RMS = 31,405 x 1.285 = 40,355A
Available Utility500,000 S.C. KVA
1000KVA Transformer,480V, 3,3.45% X, .60% R
30'  500 kcmil4 Per PhaseCopper in PVC Conduit
1600A Switch
KRPC1500SP Fuse
1 1
(Table 1.2)
(Table 3)
(Table 5)

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3 ShortCircuit Current Calculations Procedures and Methods
Ohmic Method To Fault X2 System B
Adjusted Impedanceto fault X1
400A Switch
LPSRK350SP Fuse
20'  2/02 Per PhaseCopper in PVC Conduit
225KVA Transformer,208/120V,.998%X, .666%R
R X
X = .008688 .008688
R = .001545 .001545
X = .00008 .00008
X =20' x .0327 = .000327
.000327
1000 2
R =20' x .0812 = .000812 .000812
1000 2
Total R and X (480V) = .002357 .009095
To Reflect X and R to secondary:
Xtotal =(208)2 x (.009095)
= .001708 .001708(208V) (480)2
Rtotal =(208)2 x (.002357)
= .000442 .000442 (208V) (480)2
X =(10) (.998) (.208)2
= .00192 .00192225
R =(10) (.666) (.208)2
= .00128 .00128 225
Total R and X (208V) = .001722 .003628
Ztotal per = (.001722)2 + (.003628)2 = .004015phase
IS.C. sym RMS =208 = 29,911A
3 (.004015)
X/Rratio =.003628 = 2.10.001722
Asym Factor = 1.0491 (Table 8)
IS.C. asym RMS = 29,911 x 1.0491 = 31,380A
10
OneLine Diagram Impedance Diagram
1 1
2 2
(Table 5)
(Table 1.2)

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3 ShortCircuit Current Calculations Procedures and Methods
PerUnit Method
11
3 Short Circuit Calculation PerUnit Method*The perunit method is generally used for calculating
shortcircuit currents when the electrical system is more
complex.
After establishing a oneline diagram of the system,proceed to the following calculations: **
Step 1. PUXutility =KVAbase
S.C. KVAutility
Step 2. PUXtrans =(%X)(KVAbase )
(100)(KVAtrans)
PURtrans =(%R)(KVAbase)
(100)(KVAtrans)
Step 3. PUXcomponent (cable, =(X)(KVAbase)
switches, CT, bus) (1000)(KV)2
Step 4. PURcomponent (cable, =(R)( KVAbase)
switches, CT, bus) (1000)(KV)2
Step 5. Next, total all perunit X and all perunit R in systemto point of fault.
Step 6. Determine the perunit impedance of the system by:
PUZtotal = (PURtotal)2 + (PUXtotal)2
Step 7. Calculate the symmetrical RMS shortcircuit currentat the point of fault.
IS.C. sym RMS=KVAbase
3 (KV)(PUZtotal)
Step 8. Determine the motor load. Add up the full loadmotor currents.(Whenever motor and lighting loads are
considered, such as supplied by 4 wire, 208Y/120 and
480Y/277 volt 3 phase systems, the generally accepted
procedure is to assume 50% motor load based on the full
load current rating of the transformer.)
Step 9. The symmetrical motor contribution can beapproximated by using an average multiplying factor
associated with the motors in the system. This factor varies
according to motor design and in this text may be chosen
as 4 times motor full load current for approximate
calculation purposes. To solve for the symmetrical motorcontribution:
***Isym motor contrib = (4) x (Ifull load motor)
Step 10. The total symmetrical shortcircuit rms current iscalculated as:
Itotal S.C. sym RMS = (IS.C. sym RMS) + (Isym motor contrib)
Step 11. Determine X/R ratio of the system to the point offault.
X/Rratio =
PUXtotalPURtotal
Step 12. From Table 8, Column Mm, obtain the asymmetricalfactor corresponding to the X/R ratio determined in Step
11. This multiplier will provide the worst case asymmetry
occurring in the first 1/2 cycle. When the average 3phase
multiplier is desired use column Ma.
Step 13. The asymmetrical RMS shortcircuit current can
be calculated as:
IS.C. asym RMS = (IS.C. sym RMS) x (Asym Factor)
Step 14. The shortcircuit current that the motor load can
contribute is an asymmetrical current usually approximated
as being equal to the locked rotor current of the motor.***
As a close approximation with a margin of safety use:
***Iasym motor contrib = (5) x (Ifull load motor)
Step 15. The total asymmetrical shortcircuit RMS current
is calculated as:
ItotalS.C. asym RMS = (IS.C. asym RMS) + (Iasym motor contrib)
* The base KVA used throughout this text will be 10,000 KVA.** As in the ohmic method procedure, all ohmic values are singlephase distance one way, later compensated for in the three phase shortcircuit formula by the
factor, 3. (See Step 7.) UL Listed transformers 25KVA and larger have a 10% impedance tolerance. Short circuit amperes can be affected by this tolerance. Only perunit X is considered in this procedure since utility X/R ratio is usually quite high. For more finite details obtain perunit R of utility source.
*** A more exact determination depends upon the subtransient reactance of the motors in question and associated circuit impedances. A less conservativemethod would involve the total motor circuit impedance to a common bus (sometimes referred to as a zero reactance bus).
Arithmetical addition results in conservative values of fault current. More finite values involve vectorial addition of the currents.

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*
10,000 KVA Base
PUR PUX
PUX =10,000 = 0.0001 0.0001
100,000,000
PUX =(3.45) (10,000)
= 0.2300 0.2300(100) (1500)
PUR =(.56) (10,000)
= 0.0373 0.0373 (100) (1500)
(25')x
(.0379)x (10,000)
PUX =(1000) (6)
= 0.00685 0.00685(1000) (.480)2
(25') x (.0244) x (10,000)(1000) (6)
PUR =(1000) (.480)2
= 0.0044 0.0044
PUX =(.00005) (10,000)
= 0.00217 0.00217(1000) (.480)2
Total PUR and PUX = 0.0417 0.2391
PUZtotal = (0.0417)2 + (0.2391)2 = .2430
IS.C. sym RMS =10,000 = 49,489A
3 (.480)(.2430)
Isym motor contrib = 4 x 1804 = 7,216A
Itotal S.C. sym RMS = 49,489 + 7,216 = 56,705A(fault X1)
X/Rratio =.2391 = 5.73.0417
Asym Factor = 1.294 (Table 8)
IS.C. asym RMS = 49,489 x 1.294 = 64,039A
Iasym motor contrib = 5 x 1804 = 9,020A(100% motor load)
Itotal S.C. asym RMS = 64,039 + 9,020 = 73,059A(fault X1)
3 ShortCircuit Current Calculations Procedures and Methods
PerUnit Method To Fault X1 System A
12
Available UtilityS.C. MVA 100,000
1500 KVA Transformer,480V, 3,3.5%Z, 3.45%X, .56%R
If.l. trans = 1804A
25  500kcmil6 Per PhaseService EntranceConductors in Steel Conduit
2000A Switch
KRPC2000SP Fuse
MM
Impedance DiagramOneLine Diagram
Note: See Per Unit Method Procedure for Formulas.Actual motor contribution will be somewhat smaller than calculateddue to impedance of the feeder cable.
1 1

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M M
3 ShortCircuit Current Calculations  Procedures and Methods
PerUnit Method To Fault X2 System A
13
Fault X1
400A Switch
LPSRK400SP Fuse
50  500kcmilFeeder Cable inSteel Conduit
Motor Contribution
OneLine Diagram Impedance Diagram10,000 KVA Base
PUR PUX
PUX = .2391 .2391PUR = .0417 .0417
PUX =(.00008) (10,000)
= .0034 .0034(1000) (.480)2
50 x (.0379) x (10,000)
PUX =1000
= .0822 .0822(1000) (.480)2
50 x (.0244) x (10,000)
PUR =1000 = .0529 .0529
(1000) (.480)2
Total PUR and PUX = .0946 .3247
PUZtotal = (.0946)2
+ (.3247)2
= 0.3380
IS.C. sym RMS =10,000 = 35,621A
3 (.480)(.3380)
Isym motor contrib = 4 x 1804 = 7,216A
Itotal S.C. sym RMS = 35,621 + 7,216 = 42,837A(fault X2)
X/Rratio =.32477
= 3.43.09465
Asym Factor = 1.149 (Table 8)
IS.C. asym RMS = 1.149 x 35,621 = 40,929A
Iasym motor contrib = 5 x 1804 = 9,020A(100%motor load)
Itotal S.C. asym RMS = 40,929 + 9,020 = 49,949A(fault X2)
Adjusted Impedanceto Fault X
1
1 1
2 2

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10,000KVA BasePUR PUX
PUX =10,000 = .02 .02
500,000
PUX =(3.45) (10,000)
= .345 .345(100) (1000)
PUR =(.6) (10,000)
= .06 .06 (100) (1000)
(30')x
(.0303)
PUX =(1000) (4)
x (10,000)
= .0099 .0099(1000) (.48)2
(30') x (.0220)
PUR =(1000) (4)
x (10,000)
= .0072 .0072 (1000) (.48)2
PUX =(.00005) (10,000)
= .0022 .0022(1000) (.48)2
Total PUR and PUX = .0672 .3771
PUZtotal
= (.0672)2 + (.3771)2 = .383
IS.C. sym RMS =10,000 = 31,405A
3 (.48)(.383)
X/Rratio =.3771 = 5.62.0672
Asym Factor = 1.285 (Table 8)
IS.C.asym RMS = 31,405 x 1.285 = 40,355A
14
3 ShortCircuit Current Calculations Procedures and Methods
PerUnit Method To Fault X1 System B
Available UtilityS.C. KVA 500,000
1000 KVA Transformer,480V, 33.45%X, .60%R
30'  500kcmil
4 Per PhaseCopper in PVC Conduit
1600A Switch
KRPC1500SP Fuse
OneLine Diagram Impedance Diagram
1 1

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3 ShortCircuit Current Calculations Procedures and Methods
PerUnit Method To Fault X2 System B
15
OneLine Diagram Impedance Diagram10,000 KVA
PUR PUX
X1 = .3771 .3771
R1 = .0672 .0672
PUX =(.00008) (10,000)
= .0035 .0035(1000) (.48)2
PUX =
(20')x
(.0327)x (10,000)
= .0142 .0142(1000) (2)
(1000) (.48)2
(20')x
(.0812)x (10,000)
PUR =(1000) (2)
= .0352 .0352 (1000) (.48)2
PUX =(.998) (10,000)
= .4435 .4435(100) (225)
PUR =(.666) (10,000)
= .296 .296 (100) (225)
Total PUR and PUX .3984 .8383
PUZtotal = (.3984)2 + (.8383)2 = .928
IS.C.sym RMS = 10,000 = 29,911A(3)(.208)(.928)
X/Rratio =.8383 = 2.10.3984
Asym Factor=
1.0491 (Table 8)IS.C. asym RMS = 29,911 x 1.0491 = 31,380A
Adjusted Impedance toFault X1
400A Switch
LPSRK350SP Fuse
20  2/02 Per PhaseCopper in PVC conduit
225KVA Transformer,208V, 3.998%X, .666%R
2 2
1 1

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BUSSPOWER is a Computer Software Program which
calculates three phase fault currents. It is a part of the
TRON Software Package for Power Systems Analysis.
The user inputs data which includes:
 Cable and Busway Lengths and Types Transformer Rating and Impedence Fault sources such as Utility Available and Motor
Contribution.
Following the data input phase, the program is
executed and an output report reviewed.
The following is a partial output report of System A
being studied.
TRON Software Fault Calculation Program Three Phase Fault Report
SYSTEM AFault Study Summary
Bus Record Voltage Available RMS Duties
Name LL 3 Phase Momentary
(Sym) (Asym)
X1 480 58414 77308
X2 480 44847 53111
The following is a partial output report of the
distribution System B.
SYSTEM BFault Study Summary
Bus Record Voltage Available RMS Duties
Name LL 3 Phase Momentary
(Sym) (Asym)X1 480 31,363 40,141
X2 208 29,980 31,425
A further description of this program and its
capabilities is on the back cover of this bulletin.
3 ShortCircuit Current Calculations Procedures and Methods
TRON Computer Software Method

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The application of the pointtopoint method permits the
determination of available shortcircuit currents with a
reasonable degree of accuracy at various points for either
3 or 1 electrical distribution systems. This method can
assume unlimited primary shortcircuit current (infinite bus).
Basic PointtoPoint Calculation ProcedureStep 1. Determine the transformer full load amperes fromeither the nameplate or the following formulas:
3 Transformer If.l. =KVA x 1000ELL x 1.732
1 Transformer If.l. =KVA x 1000
ELL
Step 2. Find the transformer multiplier.
Multiplier = 100*%Z trans
Note. Transformer impedance (Z) helps to determine what the short circuitcurrent will be at the transformer secondary. Transformer impedance is
determined as follows: The transformer secondary is short circuited. Voltageis applied to the primary which causes full load current to flow in thesecondary. This applied voltage divided by the rated primary voltage is theimpedance of the transformer.Example: For a 480 volt rated primary, if 9.6 volts causes secondary full loadcurrent to flow through the shorted secondary, the transformer impedance is9.6/480 = .02 = 2%Z.In addition, UL listed transformer 25KVA and larger have a 10%impedance tolerance. Short circuit amperes can be affected by thistolerance.
Step 3. Determine the transformer letthru shortcircuitcurrent**.
IS.C. = If.l. x Multiplier
Note. Motor shortcircuit contribution, if significant, may be added to the
transformer secondary shortcircuit current value as determined in Step 3.Proceed with this adjusted figure through Steps 4, 5 and 6. A practicalestimate of motor shortcircuit contribution is to multiply the total motorcurrent in amperes by 4.
Step 4. Calculate the "f" factor.
3 Faults f = 1.732 x L x IC x ELL
1 LinetoLine (LL)2 x L x I
Faults on 1 Center f =C x ELLTapped Transformer
1 LinetoNeutral2 x L x I(LN) Faults on 1 f =C x ELNCenter Tapped Transformer
Where:L = length (feet) of circuit to the fault.C = constant from Table 6, page 27. For parallel
runs, multiply C values by the number of
conductors per phase.
I = available shortcircuit current in amperes atbeginning of circuit.
Note. The LN fault current is higher than the LL fault current at thesecondary terminals of a singlephase centertapped transformer. Theshortcircuit current available (I) for this case in Step 4 should be adjustedat the transformer terminals as follows:At LN center tapped transformer terminals,I = 1.5 x LL ShortCircuit Amperes at Transformer Terminals
At some distance from the terminals, depending upon wire size, the LN faultcurrent is lower than the LL fault current. The 1.5 multiplier is anapproximation and will theoretically vary from 1.33 to 1.67. These figures arebased on change in turns ratio between primary and secondary, infinitesource available, zero feet from terminals of transformer, and 1.2 x %X and1.5 x %R for LN vs. LL resistance and reactance values. Begin LNcalculations at transformer secondary terminals, then proceed pointtopoint.
Step 5. Calculate "M" (multiplier).
M = 11 + f
Step 6. Calculate the available shortcircuit symmetricalRMS current at the point of fault.
IS.C. sym RMS = IS.C. x M
Calculation of ShortCircuit Currentsat Second Transformer in System
Use the following procedure to calculate the level of
fault current at the secondary of a second, downstream
transformer in a system when the level of fault current at the
transformer primary is known.
Procedure for Second Transformer in System
Step 1. Calculate the "f" factor (IS.C. primary known)
3 Transformer(IS.C. primary and f =
IS.C. primary x Vprimary x 1.73 (%Z)
IS.C. secondary are 100,000 x KVA trans3 fault values)
1 Transformer(IS.C. primary andIS.C. secondary are f =
IS.C. primary x Vprimary x (%Z)
1 fault values: 100,000x KVA trans
IS.C. secondary is LL)
Step 2. Calculate "M" (multiplier).
M =1
1 + f
Step 3. Calculate the shortcircuit current at the secondaryof the transformer. (See Note under Step 3 of "Basic Point
toPoint Calculation Procedure".)
IS.C. secondary =Vprimary x M x IS.C. primary
Vsecondary
**
*
3 ShortCircuit Current Calculations Procedures and Methods
PointtoPoint Method
17
IS.C. primary
MAINTRANSFORMER
H.V. UTILITYCONNECTION
IS.C. secondary
IS.C. primary IS.C. secondary

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Available UtilityS.C. MVA 100,000
1500 KVA Transformer,480V, 3, 3.5%Z,3.45%X, 56%R
If.l. =1804A
25'  500kcmil6 Per PhaseService EntranceConductors in Steel Conduit
2000A Switch
KRPC2000SP Fuse
Fault X1
400A Switch
LPSRK400SP Fuse
50'  500 kcmilFeeder Cablein Steel Conduit
Fault X2
Motor Contribution M
3 ShortCircuit Current Calculations Procedures and Methods
PointtoPoint Method To Faults X1 & X2 System A
Fault X1
Step 1. If.l. =1500 x 1000 = 1804A480 x 1.732
Step 2. Multiplier = 100 = 28.573.5
Step 3. IS.C.= 1804 x 28.57 = 51,540A
Step 4. f =1.732 x 25 x 51,540 = 0.0349
6 x 22,185 x 480
Step 5. M =1 = .9663
1 + .0349
Step 6. IS.C.sym RMS = 51,540 x .9663 = 49,803A
IS.C.motor contrib = 4 x 1,804 = 7,216A
ItotalS.C. sym RMS = 49,803 + 7,216 = 57,019A(fault X1)
Fault X2
Step 4. Use IS.C.sym RMS @ Fault X1 to calculate "f"
f =1.732 x 50 x 49,803 = .4050
22,185 x 480
Step 5. M =1 = .7117
1 + .4050
Step 6. IS.C.sym RMS = 49,803 x .7117 = 35,445A
Isym motor contrib = 4 x 1,804 = 7,216A
Itotal S.C. sym RMS(fault X2)
= 35,445 + 7,216 = 42,661A
1
2
18
OneLine Diagram

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3 ShortCircuit Current Calculations RMS Amperes
Comparison of Results
Fault X1
Step 1. If.l. =1000 x 1000 = 1203A480 x 1.732
Step 2. Multiplier = 100 = 28.573.5
Step 3. IS.C. = 1203 x 28.57 = 34,370A
Step 4. f =1.732 x 30 x 34,370 = .0348
4 x 26,706 x 480
Step 5. M =1 = .9664
1 + .0348
Step 6. IS.C.sym RMS = 34,370 x .9664 = 33,215A
Fault X2
Step 4. f = 1.732 x 20 x 33,215 = .10492 x 11,423 x 480
Step 5. M = 1 = .9051 + .1049
Step 6. IS.C.sym RMS = 33,215 x .905 = 30,059A
Fault X2
f =30,059 x 480 x 1.732 x 1.2 = 1.333
100,000 x 225
M =1 = .4286
1 + 1.333
IS.C. sym RMS =480 x .4286 x 30,059 = 29,731A
208
3 ShortCircuit Current Calculations Procedures and Methods
PointtoPoint Method To Faults X1 & X2  System B
Available Utility500,000 S.C KVA
1000 KVA Transformer,480V, 3,3.5%Z
If.l.= 1203A
30  500 kcmil4 Per PhaseCopper in PVC Conduit
1600A Switch
KRPC1500SP Fuse
Fault X1
400A Switch
LPSRK350SP Fuse
20  2/02 Per PhaseCopper in PVC Conduit
225 KVA transformer,208V, 31.2%Z
OneLine Diagram
1
2
19
System A
Ohmic PerUnit TRON PTPSym. Asym. Sym. Asym. Sym. Asym. Sym.
X1W/O Motor 49,489 64,039 49,489 64,039 49,992 64,430 49,803
W/Motor 56,705 73,059 56,705 73,059 58,414 77,308 57,019
X2W/O Motor 35,621 40,929 35,621 40,929 36,126 41,349 35,445
W/Motor 42,837 49,949 42,837 49,949 44,847 53,111 42,661
Notes:1. OHMIC and PER UNIT methods assume 100% motor contribution at X1,
then at X2.2. TRON modeled 100% motor contribution by assuming 1500 HP load,
located at Point X2.3. PTP method added symmetrical motor contribution at X1, then at X2.
System B
Ohmic PerUnit TRON PTPSym. Asym. Sym. Asym. Sym. Asym. Sym.
X1 31,405 40,355 31,405 40,355 31,363 40,145 33,215
X2 29,911 31,380 29,911 31,380 29,980 31,425 29,731

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1 ShortCircuit Current Calculations 1 Transformer System
Procedures and Methods
Shortcircuit calculations on a singlephase center
tapped transformer system require a slightly different
procedure than 3 faults on 3 systems.
1. It is necessary that the proper impedance be used to
represent the primary system. For 3 fault calculations, asingle primary conductor impedance is only considered
from the source to the transformer connection. This is
compensated for in the 3 shortcircuit formula by
multiplying the single conductor or singlephase
impedance by 1.73.
However, for singlephase faults, a primary conductor
impedance is considered from the source to the
transformer and back to the source. This is compensated in
the calculations by multiplying the 3 primary source
impedance by two.
2. The impedance of the centertapped transformer must
be adjusted for the halfwinding (generally linetoneutral)fault condition.
The diagram at the right illustrates that during lineto
neutral faults, the full primary winding is involved but, only
the halfwinding on the secondary is involved. Therefore,
the actual transformer reactance and resistance of the half
winding condition is different than the actual transformer
reactance and resistance of the full winding condition.
Thus, adjustment to the %X and %R must be made when
considering linetoneutral faults. The adjustment multipliers
generally used for this condition are as follows:
1.5 times full winding %R on full winding basis.
1.2 times full winding %X on full winding basis.
Note: %R and %X multipliers given in Table 1.3 may be used, however,calculatios must be adjusted to indicate transformer KVA/2.
3. The impedance of the cable and twopole switches on
the system must be considered "bothways" since the
current flows to the fault and then returns to the source. For
instance, if a linetoline fault occurs 50 feet from a
transformer, then 100 feet of cable impedance must be
included in the calculation.
The calculations on the following pages illustrate 1
fault calculations on a singlephase transformer system.
Both linetoline and linetoneutral faults are considered.
Note in these examples:
a. The multiplier of 2 for some electrical components to
account for the singlephase fault current flow,
b. The halfwinding transformer %X and %R multipliers for
the linetoneutral fault situation,and
c. The KVA and voltage bases used in the perunit
calculations
PRIMARY
SECONDARY
SHORTCIRCUIT
PRIMARY
SECONDARY
SHORT CIRCUIT
L2 N L1
SHORT CIRCUIT
L1
N
L2
A
B
C
50 feet
20

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100,000 KVA3 Source
75KVA, 1 Transformer,1.22%X, .68%R
Negligible Distance
400A Switch
LPNRK400SP Fuse
25'  500kcmil
Magnetic Conduit
1 ShortCircuit Current Calculations 1 Transformer System
PerUnit Method LinetoLine Fault @ 240V Fault X1
10,000KVA Base
PUR PUX
PUX(3) =10,000 = .1100,000
PUX(1) = 2 x .1 = .2000 .2000
PUX =(1.22) (10,000)
= 1.6267 1.6267(100) (75)
PUR =(.68) (10,000)
= .9067 .9067 (100) (75)
PUX =2(.00008) (10,000)
= .0278 .0278(1000) (.240)2
25'
PUX =2 x
1000x .0379 x 10,000
= .3289 .3289(1000) (.240)2
25'
PUR=2 x
1000x .0244 x 10,000
= .2118 .2118 (1000) (.240)2
Total PUR and PUX = 1.1185 2.1834
PUZtotal = (1.1185)2 + (2.1834)2 = 2.4532
IS.C. sym RMS =10,000
= 16,984ALL @ 240V (.240) (2.4532)
Note: See "Data Section" for impedance data for the electrical components.
OneLine Diagram
1 1
21
Impedance Diagram

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*
**
1 ShortCircuit Current Calculations 1 Transformer System
PerUnit Method LinetoNeutral Fault @ 120V Fault X1
100,000 KVA3 Source
75KVA, 1 Transformer,1.22%X, .68%R
Negligible Distance
400A Switch
LPNRK400SP Fuse
25'  500kcmil
Magnetic Conduit
10,000KVA Base
PUR PUX
PUX(3) =10,000 = .1100,000
PUX(1) = 2 x .1 = .2000 .2000
PUX =(1.2) (1.22) (10,000)
= 1.952 1.952(100) (75)
PUR =(1.5) (.68) (10,000)
= 1.3600 1.3600 (100) (75)
PUX* =(.00008) (10,000)
= .0556 .0556(1000) (.120)2
25'
PUX** =2 x
1000x .0379 x 10,000
= 1.316 1.316(1000) (.120)2
25'
PUR** =2 x
1000x .0244 x 10,000
= .8472 .8472 (1000) (.120)2
Total PUR and PUX = 2.2072 3.5236
PUZtotal = (2.2072)2 + (3.5236)2 = 4.158
IS.C. sym RMS =10,000
= 20,041ALN @ 120V (.120) (4.158)
Note: See "Data Section" for impedance data for the electrical components.
The multiplier of two (2) is not applicable since on a line to neutral fault, onlyone switch pole is involved.
Assumes the neutral conductor and the line conductor are the same size.
1 1
OneLine Diagram Impedance Diagram
22

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1 ShortCircuit Current Calculations 1 Transformer System
PointtoPoint Method LinetoLine Fault @ 240V Fault X1
Available UtilityS.C.KVA 100,0003 Source
75KVA, 1 Transformer,1.22%X, .68%R1.40%Z120/240V
Negligible Distance
400A Switch
LPNRK400SP Fuse
25'  500kcmil
Magnetic Conduit
OneLine Diagram
1
Fault X1
Step 1. If.l. =75 x 1000 = 312.5A
240
Step 2. Multiplier =100 = 71.431.40
Step 3. IS.C. = 312.5 x 71.43 = 22,322A
Step 4. f = 2x 25 x 22,322 = .2096
22,185 x 240
Step 5.1 = .8267M =
1 + .2096
Step 6. IS.C. LL (X1) = 22,322 x .8267 = 18,453A
23

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*
1 Short Circuit Calculations RMS Amperes
Comparison of Results
PerUnit Method vs. PointtoPoint MethodPerUnit PTP
Method Method
X1LineLine 16,984A 18,453A
LineNeutral 20,041A 20,555A
Fault X1
Step 1. If.l. =75 x 1000 = 312.5A
240
Step 2. Multiplier =100 = 71.431.40
Step 3. IS.C. (LL) = 312.5 x 71.43 = 22,322A
IS.C. (LN) = 22,322 x 1.5 = 33,483A
Step 4. f =2*x 25 x 22,322 x 1.5
= .628822,185 x 120
Step 5. 1 = .6139M =1 + .6288
Step 6. IS.C. LN (X1) = 33,483 x .6139 = 20,555A
Assumes the Neutral conductor and the line conductor are the same size.
Available UtilityS.C.KVA 100,0003 Source
75KVA, 1 Transformer,1.22% X, .68%R,1.40%Z120/240V
Negligible Distance
400A Switch
LPNRK400SP Fuse
25'  500kcmil
Magnetic Conduit
OneLine Diagram
1 ShortCircuit Current Calculations 1 Transformer System
PointtoPoint Method LinetoNeutral Fault @ 120V Fault X1
1
24

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*
**
Table 1.4. Impedance Data for Single Phase and Three PhaseTransformersSupplement
KVA Suggested
1 3 %Z X/R Ratio for Calculation
10 1.2 1.1
15 1.3 1.175 1.11 1.5
150 1.07 1.5
225 1.12 1.5
300 1.11 1.5
333 1.9 4.7
500 2.1 5.5
These represent actual transformer nameplate ratings taken from field
installations.
Note: UL Listed transformers 25KVA and greater have a 10% tolerance on
their impedance nameplate.
Table 2. Current Transformer Reactance DataApproximate Reactance of Current Transformers*
Reactance in Ohms for
Primary Current Various Voltage Ratings
Ratings  Amperes 6005000V 7500V 15,000V
100  200 0.0022 0.0040
250  400 0.0005 0.0008 0.0002
500  800 0.00019 0.00031 0.00007
1000  4000 0.00007 0.00007 0.00007
Note: Values given are in ohms per phase. For actual values, refer to manu
facturers' data.
This table has been reprinted from IEEE Std 2411990, IEEE Recommended
Practice for Commercial Building Power Systems, Copyright 1990 by the
Institute of Electrical and Electronics Engineers, Inc. with the permission of
the IEEE Standards Department.
Table 3. Disconnecting Switch Reactance Data(DisconnectingSwitch Approximate Reactance Data, in Ohms*)Switch Size Reactance(Amperes) (Ohms)
200 0.0001400 0.00008
600 0.00008
800 0.00007
1200 0.00007
1600 0.00005
2000 0.00005
3000 0.00004
4000 0.00004
Note: The reactance of disconnecting switches for lowvoltage circuits
(600V and below) is in the order of magnitude of 0.00008  0.00005
ohm/pole at 60 Hz for switches rated 400  4000 A, respectively.
*For actual values, refer to manufacturers data.
This table has been reprinted from IEEE Std 2411990, IEEE Recommended
Practice for Commercial Building Power Systems, Copyright 1990 by the
Institute of Electrical and Electronics Engineers, Inc. with the permission ofthe IEEE Standards Department.
1 Pole
25
Table 1.1. Transformer Impedance Data(X/R Ratio of Transformers Based on ANSI/IEEE C37.0101979)
This table has been reprinted from IEEE Std 1411986, IEEE Recommended
Practice for Electric Power Distribution for Industrial Plants, Copyright 1986
by the Institute of Electrical and Electronics Engineers, Inc with the
permission of the IEEE Standards Department.
Table 1.2. Impedance Data for Three Phase TransformersKVA %R %X %Z X/R
3.0 3.7600 1.0000 3.8907 0.265
6.0 2.7200 1.7200 3.2182 0.632
9.0 2.3100 1.1600 2.5849 0.502
15.0 2.1000 1.8200 2.7789 0.867
30.0 0.8876 1.3312 1.6000 1.5
45.0 0.9429 1.4145 1.7000 1.5
75.0 0.8876 1.3312 1.6000 1.5
112.5 0.5547 0.8321 1.0000 1.5
150.0 0.6657 0.9985 1.2000 1.5
225.0 0.6657 0.9985 1.2000 1.5
300.0 0.6657 0.9985 1.2000 1.5
500.0 0.7211 1.0816 1.3000 1.5
750.0 0.6317 3.4425 3.5000 5.451000.0 0.6048 3.4474 3.5000 5.70
1500.0 0.5617 3.4546 3.5000 6.15
2000.0 0.7457 4.9441 5.0000 6.63
2500.0 0.7457 4.9441 5.0000 6.63
Note: UL Listed transformers 25KVA and greater have a 10% tolerance on
their nameplate impedance.
Table 1.3. Impedance Data for Single Phase TransformersSuggested Normal Range Impedance Multipliers**
X/R Ratio of Percent For LinetoNeutral
kVA for Impedance (%Z)* Faults
1 Calculation for %X for%R
25.0 1.1 1.26.0 0.6 0.75
37.5 1.4 1.26.5 0.6 0.75
50.0 1.6 1.26.4 0.6 0.75
75.0 1.8 1.26.6 0.6 0.75
100.0 2.0 1.35.7 0.6 0.75
167.0 2.5 1.46.1 1.0 0.75
250.0 3.6 1.96.8 1.0 0.75
333.0 4.7 2.46.0 1.0 0.75
500.0 5.5 2.25.4 1.0 0.75
National standards do not speciify %Z for singlephase transformers. Consultmanufacturer for values to use in calculation.Based on rated current of the winding (onehalf nameplate kVA divided bysecondary linetoneutral voltage).
Note: UL Listed transformers 25 KVA and greater have a 10% tolerance ontheir impedance nameplate.
This table has been reprinted from IEEEStd 2421986 (R1991), IEEE
Recommended Practice for Protection and Coordination of Industrial and
Commercial Power Systems, Copyright 1986 by the Institute of Electrical
and Electronics Engineers, Inc. with the permission of the IEEE Standards
Department.
50
40
30
20
10
0
TypicalX/R
0.5 1 2 5 10 20 50 100 200 500 1000SelfCooled Transformer Rating in MVA
Data Section
Impedance and Reactance DataTransformers and Switches

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Data Section
Impedance & Reactance DataCircuit Breakers and Conductors
Table 5. Impedance Data  Insulated Conductors(Ohms/1000 ft. each conductor  60Hz)Size Resistance (25C) Reactance  600V  THHN
AWG or Copper Aluminum Single Conductors 1 Multiconductor
kcM Metal NonMet Metal Nonmet Mag. Nonmag. Mag Nonmag.
14 2.5700 2.5700 4.2200 4.2200 .0493 .0394 .0351 .030512 1.6200 1.6200 2.6600 2.6600 .0468 .0374 .0333 .0290
10 1.0180 1.0180 1.6700 1.6700 .0463 .0371 .0337 .0293
8 .6404 .6404 1.0500 1.0500 .0475 .0380 .0351 .0305
6 .4100 .4100 .6740 .6740 .0437 .0349 .0324 .0282
4 .2590 .2590 .4240 .4240 .0441 .0353 .0328 .0235
2 .1640 .1620 .2660 .2660 .0420 .0336 .0313 .0273
1 .1303 .1290 .2110 .2110 .0427 .0342 .0319 .0277
1/0 .1040 .1020 .1680 .1680 .0417 .0334 .0312 .0272
2/0 .0835 .0812 .1330 .1330 .0409 .0327 .0306 .0266
3/0 .0668 .0643 .1060 .1050 .0400 .0320 .0300 .0261
4/0 .0534 .0511 .0844 .0838 .0393 .0314 .0295 .0257
250 .0457 .0433 .0722 .0709 .0399 .0319 .0299 .0261
300 .0385 .0362 .0602 .0592 .0393 .0314 .0295 .0257
350 .0333 .0311 .0520 .0507 .0383 .0311 .0290 .0254
400 .0297 .0273 .0460 .0444 .0385 .0308 .0286 .0252
500 .0244 .0220 .0375 .0356 .0379 .0303 .0279 .0249
600 .0209 .0185 .0319 .0298 .0382 .0305 .0278 .0250750 .0174 .0185 .0264 .0240 .0376 .0301 .0271 .0247
1000 .0140 .0115 .0211 .0182 .0370 .0296 .0260 .0243
Note: Increased resistance of conductors in magnetic raceway is due to the effect of hysteresislosses. The increased resistance of conductors in metal nonmagnetic raceway is due to the effectof eddy current losses. The effect is essentially equal for steel and aluminum raceway. Resistancevalues are acceptable for 600 volt, 5KV and 15 KV insulated Conductors.
Size Reactance  5KV Reactance  15KV
AWG or Single Conductors 1 Multiconductor Single Conductors 1 Multiconductor
kcM Mag. Nonmag. Mag. Nonmag. Mag. Nonmag. Mag. Nonmag.
8 .0733 .0586 .0479 .0417
6 .0681 .0545 .0447 .0389 .0842 .0674 .0584 .0508
4 .0633 .0507 .0418 .0364 .0783 .0626 .0543 .0472
2 .0591 .0472 .0393 .0364 .0727 .0582 .0505 .0439
1 .0571 .0457 .0382 .0332 .0701 .0561 .0487 .0424
1/0 .0537 .0430 .0360 .0313 .0701 .0561 .0487 .0424
2/0 .0539 .0431 .0350 .0305 .0661 .0561 .0458 .0399
3/0 .0521 .0417 .0341 .0297 .0614 .0529 .0427 .03724/0 .0505 .0404 .0333 .0290 .0592 .0491 .0413 .0359
250 .0490 .0392 .0323 .0282 .0573 .0474 .0400 .0348
300 .0478 .0383 .0317 .0277 .0557 .0458 .0387 .0339
350 .0469 .0375 .0312 .0274 .0544 .0446 .0379 .0332
400 .0461 .0369 .0308 .0270 .0534 .0436 .0371 .0326
500 .0461 .0369 . 0308 .0270 .0517 .0414 .0357 .0317
600 .0439 .0351 .0296 .0261 .0516 .0414 . 0343 .0309
750 .0434 .0347 .0284 .0260 .0500 .0413 .0328 .0301
1000 .0421 .0337 .0272 .0255 .0487 .0385 .0311 .0291
These are only representative figures. Reactance is affected by cable insulation type, shielding,
conductor outside diameter, conductor spacing in 3 conductor cable, etc. In commercial buildings
meduim voltage impedances normally do not affect the short circuit calculations significantly.
This table has been reprinted from IEEE Std 2411990, IEEE Recommended Practice for Commercial
Building Power Systems, copyright 1990 by the Institute of Electrical and Electronics Engineers,
Inc. with the permission of the IEEE Standards Department.
*
Table 4. Circuit Breaker Reactance Data(a) Reactance of LowVoltage Power Circuit Breakers
CircuitBreaker
Interrupting CircuitBreaker
Rating Rating Reactance(amperes) (amperes) (ohms)
15,000 15  35 0.04
and 50  100 0.004
25,000 125  225 0.001
250  600 0.0002
50,000 200  800 0.0002
1000  1600 0.00007
75,000 2000  3000 0.00008
100,000 4000 0.00008
(b)Typical Molded Case Circuit Breaker Impedances
MoldedCase
CircuitBreaker
Rating Resistance Reactance
(amperes) (ohms) (ohms)
20 0.00700 Negligible
40 0.00240 Negligible
100 0.00200 0.00070
225 0.00035 0.00020
400 0.00031 0.00039
600 0.00007 0.00017
Notes:
(1) Due to the method of rating lowvoltage powercircuit breakers, the reactance of the circuit breakerwhich is to interrupt the fault is not included incalculating fault current.
(2) Above 600 amperes the reactance of molded casecircuit breakers are similar to those given in (a)
For actual values, refer to manufacturers data.
This table has been reprinted from IEEE Std 2411990,
IEEE Recommended Practice for Commercial Building
Power Systems, copyright 1990 by the Institute of
Electrical and Electronics Engineers, Inc. with the
permission of the IEEE Standards Department.
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Data Section
"C" Values for Conductors and Busway
Table 6. C Values for Conductors and BuswayCopper
AWG Three Single Conductors ThreeConductor Cable
or Conduit Conduit
kcmil Steel Nonmagnetic Steel Nonmagnetic
600V 5KV 15KV 600V 5KV 15KV 600V 5KV 15KV 600V 5KV 15KV
14 389 389 389 389 389 389 389 389 389 389 389 389
12 617 617 617 617 617 617 617 617 617 617 617 617
10 981 981 981 981 981 981 981 981 981 981 981 981
8 1557 1551 1557 1558 1555 1558 1559 1557 1559 1559 1558 1559
6 2425 2406 2389 2430 2417 2406 2431 2424 2414 2433 2428 2420
4 3806 3750 3695 3825 3789 3752 3830 3811 3778 3837 3823 3798
3 4760 4760 4760 4802 4802 4802 4760 4790 4760 4802 4802 4802
2 5906 5736 5574 6044 5926 5809 5989 5929 5827 6087 6022 5957
1 7292 7029 6758 7493 7306 7108 7454 7364 7188 7579 7507 7364
1/0 8924 8543 7973 9317 9033 8590 9209 9086 8707 9472 9372 9052
2/0 10755 10061 9389 11423 10877 10318 11244 11045 10500 11703 11528 11052
3/0 12843 11804 11021 13923 13048 12360 13656 13333 12613 14410 14118 13461
4/0 15082 13605 12542 16673 15351 14347 16391 15890 14813 17482 17019 16012
250 16483 14924 13643 18593 17120 15865 18310 17850 16465 19779 19352 18001
300 18176 16292 14768 20867 18975 17408 20617 20051 18318 22524 21938 20163
350 19703 17385 15678 22736 20526 18672 19557 21914 19821 22736 24126 21982
400 20565 18235 16365 24296 21786 19731 24253 23371 21042 26915 26044 23517
500 22185 19172 17492 26706 23277 21329 26980 25449 23125 30028 28712 25916
600 22965 20567 47962 28033 25203 22097 28752 27974 24896 32236 31258 27766
750 24136 21386 18888 28303 25430 22690 31050 30024 26932 32404 31338 28303
1000 25278 22539 19923 31490 28083 24887 33864 32688 29320 37197 35748 31959
Aluminum
14 236 236 236 236 236 236 236 236 236 236 236 236
12 375 375 375 375 375 375 375 375 375 375 375 375
10 598 598 598 598 598 598 598 598 598 598 598 598
8 951 950 951 951 950 951 951 951 951 951 951 951
6 1480 1476 1472 1481 1478 1476 1481 1480 1478 1482 1481 1479
4 2345 2332 2319 2350 2341 2333 2351 2347 2339 2353 2349 2344
3 2948 2948 2948 2958 2958 2958 2948 2956 2948 2958 2958 2958
2 3713 3669 3626 3729 3701 3672 3733 3719 3693 3739 3724 3709
1 4645 4574 4497 4678 4631 4580 4686 4663 4617 4699 4681 4646
1/0 5777 5669 5493 5838 5766 5645 5852 5820 5717 5875 5851 5771
2/0 7186 6968 6733 7301 7152 6986 7327 7271 7109 7372 7328 7201
3/0 8826 8466 8163 9110 8851 8627 9077 8980 8750 9242 9164 8977
4/0 10740 10167 9700 11174 10749 10386 11184 11021 10642 11408 11277 10968
250 12122 11460 10848 12862 12343 11847 12796 12636 12115 13236 13105 12661
300 13909 13009 12192 14922 14182 13491 14916 14698 13973 15494 15299 14658
350 15484 14280 13288 16812 15857 14954 15413 16490 15540 16812 17351 16500
400 16670 15355 14188 18505 17321 16233 18461 18063 16921 19587 19243 18154
500 18755 16827 15657 21390 19503 18314 21394 20606 19314 22987 22381 20978
600 20093 18427 16484 23451 21718 19635 23633 23195 21348 25750 25243 23294
750 21766 19685 17686 23491 21769 19976 26431 25789 23750 25682 25141 23491
1000 23477 21235 19005 28778 26109 23482 29864 29049 26608 32938 31919 29135
Note: These values are equal to one over the impedance per foot for impedances found in Table 5, Page 26.
Ampacity Busway
PlugIn Feeder High Impedance
Copper Aluminum Copper Aluminum Copper
225 28700 23000 18700 12000
400 38900 34700 23900 21300
600 41000 38300 36500 31300
800 46100 57500 49300 44100
1000 69400 89300 62900 56200 15600
1200 94300 97100 76900 69900 16100
1350 119000 104200 90100 84000 17500
1600 129900 120500 101000 90900 19200
2000 142900 135100 134200 125000 20400
2500 143800 156300 180500 166700 21700
3000 144900 175400 204100 188700 23800
4000 277800 256400
Note: These values are equal to one over the impedance per foot for
impedances in Table 7, Page 28.
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Data Section
Busway Impedance Data
Table 7. Busway Impedance Data (Ohms per 1000 Feet LinetoNeutral, 60 Cycles)PlugIn Busway
Copper Bus Bars Aluminum Bus Bars
Ampere Rating Resistance Reactance Impedance Resistance Reactance Impedance
225 0.0262 0.0229 0.0348 0.0398 0.0173 0.0434
400 0.0136 0.0218 0.0257 0.0189 0.0216 0.0288
600 0.0113 0.0216 0.0244 0.0179 0.0190 0.0261
800 0.0105 0.0190 0.0217 0.0120 0.0126 0.0174
1000 0.0071 0.0126 0.0144 0.0080 0.0080 0.0112
1200 0.0055 0.0091 0.0106 0.0072 0.0074 0.0103
1350 0.0040 0.0072 0.0084 0.0065 0.0070 0.0096
1600 0.0036 0.0068 0.0077 0.0055 0.0062 0.0083
2000 0.0033 0.0062 0.0070 0.0054 0.0049 0.0074
2500 0.0032 0.0062 0.0070 0.0054 0.0034 0.0064
3000 0.0031 0.0062 0.0069 0.0054 0.0018 0.0057
4000 0.0030 0.0062 0.0069
5000 0.0020 0.0039 0.0044
LowImpedance Feeder Busway
225 0.0425 0.0323 0.0534 0.0767 0.0323 0.0832
400 0.0291 0.0301 0.0419 0.0378 0.0280 0.0470
600 0.0215 0.0170 0.0274 0.0305 0.0099 0.0320
800 0.0178 0.0099 0.0203 0.0212 0.0081 0.0227
1000 0.0136 0.0082 0.0159 0.0166 0.0065 0.0178
1200 0.0110 0.0070 0.0130 0.0133 0.0053 0.0143
1350 0.0090 0.0065 0.0111 0.0110 0.0045 0.0119
1600 0.0083 0.0053 0.0099 0.0105 0.0034 0.0110
2000 0.0067 0.0032 0.0074 0.0075 0.0031 0.0080
2500 0.0045 0.0032 0.0055 0.0055 0.0023 0.0060
3000 0.0041 0.0027 0.0049 0.0049 0.0020 0.0053
4000 0.0030 0.0020 0.0036 0.0036 0.0015 0.0039
5000 0.0023 0.0015 0.0027
The above data represents values which are a composite of those obtained by a survey of industry; values tend to be on the low side.
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Data Section
Asymmetrical Factors
Table 8. Asymmetrical FactorsRatio to Symmetrical RMS Amperes
Short Circuit Short Maximum 1 phase Maximum 1 phase Average 3 phase
Power Factor, Circuit Instantaneous RMS Amperes at RMS Amperes at
Percent* X/R Ratio Peak Amperes Mp 1/2 Cycle Mm 1/2 Cycle Ma*
(Asym.Factor)*
0 2.828 1.732 1.394
1 100.00 2.785 1.697 1.374
2 49.993 2.743 1.662 1.354
3 33.322 2.702 1.630 1.336
4 24.979 2.663 1.599 1.318
5 19.974 2.625 1.569 1.302
6 16.623 2.589 1.540 1.286
7 14.251 2.554 1.512 1.271
8 13.460 2.520 1.486 1.256
9 11.066 2.487 1.461 1.242
10 9.9301 2.455 1.437 1.229
11 9.0354 2.424 1.413 1.216
12 8.2733 2.394 1.391 1.204
13 7.6271 2.364 1.370 1.193
14 7.0721 2.336 1.350 1.182
15 6.5912 2.309 1.331 1.172
16 6.1695 2.282 1.312 1.162
17 5.7947 2.256 1.295 1.152
18 5.4649 2.231 1.278 1.144
19 5.16672 2.207 1.278 1.135
20 4.8990 2.183 1.247 1.127
21 4.6557 2.160 1.232 1.119
22 4.4341 2.138 1.219 1.112
23 4.2313 2.110 1.205 1.105
24 4.0450 2.095 1.193 1.099
25 3.8730 2.074 1.181 1.092
26 3.7138 2.054 1.170 1.087
27 3.5661 2.034 1.159 1.081
28 3.4286 2.015 1.149 1.076
29 3.3001 1.996 1.139 1.071
30 3.1798 1.978 1.130 1.064
31 3.0669 1.960 1.122 1.062
32 2.9608 1.943 1.113 1.057
33 2.8606 1.926 1.106 1.057
34 2.7660 1.910 1.098 1.050
35 2.6764 1.894 1.091 1.046
36 2.5916 1.878 1.085 1.043
37 2.5109 1.863 1.079 1.040
38 2.4341 1.848 1.073 1.037
39 2.3611 1.833 1.068 1.034
40 2.2913 1.819 1.062 1.031
41 2.2246 1.805 1.058 1.029
42 2.1608 1.791 1.053 1.027
43 2.0996 1.778 1.049 1.024
44 2.0409 1.765 1.045 1.023
45 1.9845 1.753 1.041 1.021
46 1.9303 1.740 1.038 1.019
47 1.8780 1.728 1.035 1.017
48 1.8277 1.716 1.032 1.016
49 1.7791 1.705 1.029 1.014
50 1.7321 1.694 1.026 1.013
55 1.5185 1.641 1.016 1.008
60 1.3333 1.594 1.009 1.004
65 1.1691 1.517 1.005 1.001
70 1.0202 1.517 1.002 1.001
75 0.8819 1.486 1.0008 1.0004
80 0.7500 1.460 1.0002 1.0001
85 0.6198 1.439 1.00004 1.00002
100 0.0000 1.414 1.00000 1.00000
*Reprinted by permission of National Electrical Manufacturer's Association from
NEMA Publication AB1, 1986, copyright 1986 by NEMA.
Selective Coordination (Blackout Prevention)Having determined the faults that must be
interrupted, the next step is to specify Protective
Devices that will provide a Selectively Coordinated
System with proper Interrupting Ratings.
Such a system assures safety and reliability
under all service conditions and prevents needless
interruption of service on circuits other than the one
on which a fault occurs.
The topic of Selectivity will be Discussed in the
next Handbook, EDP II.
Component Protection (Equipment Damage Prevention)Proper protection of electrical equipment
requires that fault current levels be known. The
characteristics and letthrough values of the
overcurrent device must be known, and compared
to the equipment withstand ratings. This topic ofComponent Protection is discussed in the third
Handbook, EDP III.